What is the surface area of the solid created by revolving #f(x)=e^(x^2+x-1)/(x+1)# over #x in [0,1]# around the x-axis?

Question

Cameron Alexander · Accepted Answer

# ~~1.45099#

The Volume of Revolution about #Ox# is given by:

# V= int_(x=a)^(x=b) \ pi y^2 \ dx #

So for for this problem:

# V= int_0^1 \ pi (e^(x^2+x-1)/(x+1))^2 \ dx #

There is no elementary anti-derivative. The solution can be found numerically as #~~1.45099#

Christopher Agresta · Answer

undefined To find the surface area of the solid generated by revolving the function $ f(x) = \frac{e^{x^2+x-1}}{x+1} $ over the interval $[0,1]$ around the x-axis, we use the formula for surface area of a solid of revolution:

$$ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx $$

Where $ f'(x) $ denotes the derivative of $ f(x) $ with respect to $ x $, and $ a $ and $ b $ are the limits of integration.

First, we find $ f'(x) $ by differentiating $ f(x) $:

$$ f'(x) = \frac{(x^2 + 3x)e^{x^2+x-1} - e^{x^2+x-1}}{(x+1)^2} $$

Next, we substitute $ f(x) $ and $ f'(x) $ into the formula and integrate over the interval $[0,1]$:

$$ S = 2\pi \int_{0}^{1} \frac{e^{x^2+x-1}}{x+1} \sqrt{1 + \left( \frac{(x^2 + 3x)e^{x^2+x-1} - e^{x^2+x-1}}{(x+1)^2} \right)^2} \, dx $$

This integral needs to be computed numerically using appropriate numerical methods.